Question

$\rm \red{ \displaystyle\text{ \rm}\rm \sum_{k=2}^{n-1} \int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2 k+1} x \cos ^{k} x+\sin ^{k} x \cos ^{2 k+1} x}{\sin ^{3 k+3} x+\cos ^{3 k+3} x}}$​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\rm{\sum^{n-1}_{k=2}\int^{\frac{\pi}{4}}_{0}\dfrac{sin^{2k+1}(x)\cdot\,cos^{k}(x)+sin^{k}(x)\cdot\,cos^{2k+1}(x)}{sin^{3k+3}(x)+cos^{3k+3}(x)}\,dx}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\int^{\frac{\pi}{4}}_{0}\dfrac{sin^{k}(x)\,cos^{k}(x)\cdot\left\{sin^{k+1}(x)+cos^{k+1}(x)\right\}}{\left(sin^{k+1}(x)\right)^{3}+\left(cos^{k+1}(x)\right)^{3}}\,dx}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\int^{\frac{\pi}{4}}_{0}\dfrac{sin^{k}(x)\,cos^{k}(x)\cdot\left\{sin^{k+1}(x)+cos^{k+1}(x)\right\}}{\left\{sin^{k+1}(x)+cos^{k+1}(x)\right\}\left\{sin^{2k+2}(x)-sin^{k+1}(x)\,\cdot\,cos^{k+1}(x)+cos^{2k+2}(x)\right\}}\,dx}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\int^{\frac{\pi}{4}}_{0}\dfrac{sin^{k}(x)\,cos^{k}(x)}{sin^{2k+2}(x)-sin^{k+1}(x)\,\cdot\,cos^{k+1}(x)+cos^{2k+2}(x)}\,dx}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\int^{\frac{\pi}{4}}_{0}\dfrac{sin^{k}(x)\,cos^{k}(x)}{cos^{2k+2}(x)\left\{tan^{2k+2}(x)-tan^{k+1}(x)+1\right\}}\,dx}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\int^{\frac{\pi}{4}}_{0}\dfrac{\dfrac{sin^{k}(x)\,cos^{k}(x)}{cos^{2k+2}(x)}}{\left\{tan^{k+1}(x)\right\}^{2}-tan^{k+1}(x)+1}\,dx}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\int^{\frac{\pi}{4}}_{0}\dfrac{\dfrac{sin^{k}(x)\,cos^{k}(x)}{cos^{2k}(x)\cdot\,cos^{2}(x)}}{\left\{tan^{k+1}(x)\right\}^{2}-tan^{k+1}(x)+1}\,dx}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\int^{\frac{\pi}{4}}_{0}\dfrac{\dfrac{sin^{k}(x)}{cos^{k}(x)}\cdot\dfrac{1}{cos^{2}(x)}}{\left\{tan^{k+1}(x)\right\}^{2}-tan^{k+1}(x)+1}\,dx}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\int^{\frac{\pi}{4}}_{0}\dfrac{tan^{k}(x)\cdot\,sec^{2}(x)}{\left\{tan^{k+1}(x)\right\}^{2}-tan^{k+1}(x)+1}\,dx}$

$\bf{Put\,\,\,\,tan^{k+1}(x)=t}$

$\bf{\longmapsto\,\,\,(k+1)\,tan^{k}(x)\cdot\,sec^{2}(x)\,dx=dt}$

$\bf{\longmapsto\,\,\,tan^{k}(x)\cdot\,sec^{2}(x)\,dx=\dfrac{dt}{k+1}}$

Limits of integration become 0 to 1,

So,

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\int^{1}_{0}\dfrac{dt}{{t}^{2}-t+1}}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\int^{1}_{0}\dfrac{dt}{{t}^{2}-2\cdot\dfrac{1}{2}\cdot\,t+\dfrac{1}{4}+\dfrac{3}{4}}}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\int^{1}_{0}\dfrac{dt}{{t}^{2}-2\cdot\dfrac{1}{2}\cdot\,t+\left(\dfrac{1}{2}\right)^{2}+\left(\dfrac{\sqrt{3}}{2}\right)^{2}}}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\int^{1}_{0}\dfrac{dt}{\left(t-\dfrac{1}{2}\right)^{2}+\left(\dfrac{\sqrt{3}}{2}\right)^{2}}}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\cdot\dfrac{2}{\sqrt{3}}\cdot\left[tan^{-1}\left(\dfrac{t-\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)\right]^{1}_{0}}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\cdot\dfrac{2}{\sqrt{3}}\cdot\left[tan^{-1}\left(\dfrac{2t-1}{\sqrt{3}}\right)\right]^{1}_{0}}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\cdot\dfrac{2}{\sqrt{3}}\cdot\left[tan^{-1}\left(\dfrac{2-1}{\sqrt{3}}\right)-tan^{-1}\left(\dfrac{-1}{\sqrt{3}}\right)\right]}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\cdot\dfrac{2}{\sqrt{3}}\cdot\left[tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)+tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\right]}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\cdot\dfrac{2}{\sqrt{3}}\cdot\left[2\,tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\right]}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\cdot\dfrac{2}{\sqrt{3}}\cdot\left[2\times\dfrac{\pi}{6}\right]}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\cdot\dfrac{2}{\sqrt{3}}\cdot\dfrac{\pi}{3}}$

$\displaystyle=\rm{\sum^{n-1}_{k=2}\dfrac{1}{k+1}\cdot\dfrac{2\pi}{3\sqrt{3}}}$

$\displaystyle=\rm{\dfrac{2\pi}{3\sqrt{3}}\cdot\sum^{n-1}_{k=2}\dfrac{1}{k+1}}$

$\displaystyle=\rm{\dfrac{2\pi}{3\sqrt{3}}\cdot\sum^{n-1}_{k=2}\dfrac{1}{k+1}}$